In a uniform cross-section bar, the load always be divided ( shared between the bar or beam). So, if there is X number of bars holding the external weight then bar weight is equals to X/2

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I assume we are talking about longitudinal elongation under tensile / compressive force -  equal to weight of rod.

Strain =  stress / Y
ΔL / Length = W / Area Y

ΔL = W L / A Y

If the external load is Same as W,  then total tensile/compressive force is 2 W.
Then ΔL will be twice to that of ΔL under its own weight = 2 ΔL.

So elongation due to external load W is  ΔL and elongation due to its own weight W is ΔL.

Their ratio is 1 : 1

1 5 1